Question

Let $f$ be a differentiable function on $R$ and satisfying the integral equation $\underset{0}{\overset{x}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}t.f(x-t)dt=-1+e^{-x}$ for all $x\in R$, then

• A. $f\left(0\right)+f^{\prime }\left(0\right)=1$
• B. $f\left(2\right)=e^{-2}$
• C. $f^{\prime }\left(0\right)=1$
• D. $f^{\prime }\left(0\right)=2$

Answer 1

Answer: (A), (B), (D)

$f^{\prime }\left(0\right)=2$

Given equation is
$\underset{0}{\overset{x}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}tf(x-t)dt=-1+e^{-x}$
$\Rightarrow \underset{0}{\overset{x}{\int }}f\left(t\right)dt+\underset{0}{\overset{x}{\int }}(x+0-t)f\{x-(x+0-t\left)\right\}dt=-1+e^{-x}$
$\Rightarrow \underset{0}{\overset{x}{\int }}f\left(t\right)dt+x\underset{0}{\overset{x}{\int }}f\left(t\right)dt-\underset{0}{\overset{x}{\int }}tf\left(t\right)dt=-1+e^{-x}$
Differentiating both sides w.r.t. $x$, we get
$f\left(x\right)\times 1+1\times \underset{0}{\overset{x}{\int }}f\left(t\right)dt+x\left[f\right(x)\times 1-f(0)\times 0]$
$-\left[xf\right(x\left)\frac{d}{dx}\right(x\left)\right]-0f\left(0\right)\frac{d}{dx}\left(0\right)=0-e^{-x}$
$\Rightarrow f\left(x\right)+\underset{0}{\overset{x}{\int }}f\left(t\right)dt+xf\left(x\right)-xf\left(x\right)=-e^{-x}$ ... (1)
Put $x=0$ in (1), we get
$f\left(0\right)+0=-e^0=-1\Rightarrow f\left(0\right)=-1$
Again diff. both sides w.r.t. $x$, we get
$f^{\prime }\left(x\right)+\left[f\right(x)\times 1-f(0)\times 0]=e^{-x}$
$\Rightarrow e^x\left[f\right(x)+f^{\prime }(x\left)\right]=\varphi$
$\Rightarrow \frac{d}{dx}\left[e^{x}f\right(x\left)\right]=\varphi$
Integrating, we get $e^{x}f\left(x\right)=x+c$ ... (2)
Put $x=0$ in (2) $e^{0}f\left(0\right)=0+c\Rightarrow c=1\times (-1)=-1$
$\therefore$ From (2),
$f\left(x\right)=(x-1)e^{-x}$
$\therefore f\left(2\right)=(2-1)e^{-2}=e^{-2}$
$f^{\prime }\left(x\right)=-(x-1)e^{-x}+e^{-x}$
$\Rightarrow f^{\prime }\left(0\right)=-(0-1)e^0+e^0=2$
$\therefore f\left(0\right)+f^{\prime }\left(0\right)=-1+2=1$